1. **State the problem:** Differentiate the function $$f(x) = \frac{-10x+3}{-8x-3}$$ with respect to $$x$$. 2. **Formula used:** For a function $$f(x) = \frac{u(x)}{v(x)}$$, the derivative is given by the quotient rule: $$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$ where $$u(x) = -10x + 3$$ and $$v(x) = -8x - 3$$. 3. **Find derivatives of numerator and denominator:** $$ u'(x) = \frac{d}{dx}(-10x + 3) = -10$$ $$v'(x) = \frac{d}{dx}(-8x - 3) = -8$$ 4. **Apply quotient rule:** $$ f'(x) = \frac{(-10)(-8x - 3) - (-10x + 3)(-8)}{(-8x - 3)^2} $$ 5. **Simplify numerator:** $$ (-10)(-8x - 3) = 80x + 30$$ $$ (-10x + 3)(-8) = 80x - 24$$ So numerator becomes: $$ 80x + 30 - (80x - 24) = 80x + 30 - 80x + 24 = 54$$ 6. **Final derivative:** $$ f'(x) = \frac{54}{(-8x - 3)^2} $$ This is the derivative of the given function.